why i get this error Size vector must be a row vector with real elements.

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imola
imola 2015-6-9
评论： Walter Roberson 2015-6-17
Dear,
i have test this code to solve mixed integer problem but it give error her in this step
CoefMatZX = [repmat(-CoefZ,p,1),-U(1:p,:)*A; zeros(q,k+1),-V(1:q,:)*A];
it says error Size vector must be a row vector with real elements, that's because k is complex number and when i take
zeros(q,real(k)+1)
the loop doesn't stop, i don't know why.
%%Bender’s decompostion
% Application benders Decomposition algorithm to solve mixed- integer programming problem
% [OptX,OptY,OptValue]=BendersDecomposition(C,D,A,B,b)
% : To solve programming problems of the form :
% min C*x+D*y
% s.t. A*x+B*y>=b; x,0-1??variable,y>=0
% Iterative stop determination threshold value:LB/UB More than epsilon Can be flexibly adjusted according %to the complexity of the problem
function [OptX,OptY,OptValue]=BendersDecomposition(C,D,A,B,b)
C=[-1,-4];D=[-2,-3];A=[1 -3;-1 -3];B=[1 -2;-1 -1];b=[-2;-3];
epsilon = 0.99; %Stop iteration.
[nRowA,nColA] = size(A);
[nRowB,nColB] = size(B);
%step 1: Initialization
x0 = zeros(nColA,1);%The initial value
LB = -1e10;
UB = inf;
p = 0;
q = 0;
U = zeros(100,nRowA);
V = zeros(100,nRowA);
options = optimset('LargeScale', 'off', 'Simplex', 'on');
OptionsBint = optimset('MaxRLPIter',100000,'NodeSearchStrategy','bn',...
'MaxTime',50000);
while LB/UB<epsilon
%step 2: Subproblems
[UorV,fval,exitflag] = linprog((A*x0-b),B',D',[],[],zeros(nRowB,1),...
inf(nRowB,1),[],options)
if exitflag == 1 % Get extreme point
p = p+1;
U(p,:) = UorV'; % Extreme point
UB = C*x0 + U(p,:)*(b-A*x0);
elseif exitflag == -3 % Unbounded
q = q+1
V(q,:) = UorV'/1e16 % Pole direction
V(q,V(q,:)<0.0001) = 0;
else
msgbox('No feasible solution!')
return;
end
%step 3: Solving the main problem
if isinf(UB)
k = 17
else
k = ceil(log2(UB+1))-1 %k is complex number??According to 2^(k+1)-1 > UB Determines the minimum
%value of k
end
CoefZ =2.^(0:k) %z0,, The coefficient of vector zk
f = [CoefZ';zeros(nColA,1)] % Main objective function coefficient vector of
CoefMatZX = [repmat(-CoefZ,p,1),-U(1:p,:)*A; zeros(q,k+1),-V(1:q,:)*A];
bZX = [-(repmat(C*x0,p,1) + U(1:p,:)*b);-V(1:q,:)*b];
[OptZX,minZ,ExitflagBint] = bintprog(f,CoefMatZX,bZX,[],[],[],...
OptionsBint);
if ExitflagBint ==1
LB = minZ;
x0 = OptZX(k+2:end);
else
break;
end
LB,UB %#ok<NOPRT>
end
OptX = x0;
options = optimset('LargeScale', 'off', 'Simplex', 'on');
[OptY,OptValue] = linprog(D',-B,A*OptX-b,[],[],zeros(nColB,1),inf(nColB,1),...
[],options);
OptValue = C*OptX+D*OptY;
if true
% code
end
if true
% code
end
I need to run this code because there is no other one in matlab and it is useful to my work. thanks for any help.
regards,
Imola
  2 个评论
Alka Nair
Alka Nair 2015-6-17
Hi, real(k) does not guarantee that it is an integer. Try explicitly making the dimension an integer value. Thank, Alka
Walter Roberson
Walter Roberson 2015-6-17
In order for k to become complex, UB would have to be less than -1. Between -1 (inclusive) and 0 (inclusive) for UB, k would become negative and that would give problems. I do not understand the linear algebra well enough to see any reason why k < -1 should not be possible.

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